The Consumer Reports National Research Center conducted a telephone survey of 2,000 adults to learn about the major economic concerns for the future. The survey results showed that 1,620 of the respondents think the future health of Social Security is a major economic concern.
a. What is the point estimate of the population
proportion of adults who think the future health of Social Security
is a major economic concern (to 3 decimals)?
b. At 90% confidence, what is the margin of
error (to 4 decimals)? Use critical value with three decimal
places.
c. Develop a 90% confidence interval for the population proportion of adults who think the future health of Social Security is a major economic concern (to 3 decimals). Use critical value with three decimal places.
(___ , ___)
d. Develop a 95% confidence interval for this population proportion (to 4 decimals).
(___ , ___)
e. Develop a 99% confidence interval for this population proportion (to 4 decimals).
(___ , ___)
Solution :
Given that,
n = 2000
x = 1620
a) Point estimate = sample proportion = = x / n = 1620 / 2000 = 0.810
1 - = 1 - 0.810 = 0.190
b) At 90% confidence level
= 1 - 90%
=1 - 0.90 =0.10
/2
= 0.05
Z/2
= Z0.05 = 1.645
Margin of error = E = Z / 2 * (( * (1 - )) / n)
= 1.645 (((0.810 * 0.190) / 2000)
= 0.0144
c) A 90% confidence interval for population proportion p is ,
± E
= 0.810 ± 0.0144
= ( 0.7956, 0.8244 )
d) At 95% confidence level
= 1 - 95%
=1 - 0.95 =0.05
/2
= 0.025
Z/2
= Z0.025 = 1.960
Margin of error = E = Z / 2 * (( * (1 - )) / n)
= 1.96 (((0.810 * 0.190) / 2000)
= 0.0172
A 95% confidence interval for population proportion p is ,
± E
= 0.810 ± 0.0172
= ( 0.7928, 0.8272 )
e) At 99% confidence level
= 1 - 99%
=1 - 0.99 =0.01
/2
= 0.005
Z/2
= Z0.005 = 2.576
Margin of error = E = Z / 2 * (( * (1 - )) / n)
= 2.576 (((0.810 * 0.190) / 2000)
= 0.0226
A 99% confidence interval for population proportion p is ,
± E
= 0.810 ± 0.0226
= ( 0.7874, 0.8326 )
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